How I can maximize this problem using the KKT method?
\(\displaystyle f(u,z,l)=\sqrt[ ]{(xz)^2+l^2}\)
with the next conditions:
\(\displaystyle \displaystyle\frac{x}{3}+\displaystyle\frac{z}{2}+l=24\)
\(\displaystyle x\geq{0}\);\(\displaystyle z\geq{0}\);\(\displaystyle l>o\)
\(\displaystyle x+z>0\)
I do the next lagrangian:
\(\displaystyle L=-(xz)^2-l^2+a(\displaystyle\frac{x}{3}+\displaystyle\frac{z}{2}+l-24)+b(-x-z)\)
and then I have the KKT conditions:
\(\displaystyle -2xz^2+\displaystyle\frac{a}{3}-b\geq{0}\) \(\displaystyle (-2xz^2+\displaystyle\frac{a}{3}-b)x=0\)
\(\displaystyle -2x^2z+\displaystyle\frac{a}{2}-b\geq{0}\) \(\displaystyle (-2x^2z+\displaystyle\frac{a}{2}-b)z=0\)
\(\displaystyle -2l+a\leq{0}\) \(\displaystyle (-2l+a)l=0\)
\(\displaystyle \displaystyle\frac{x}{3}+\displaystyle\frac{z}{2}+l-24=0\) \(\displaystyle (\displaystyle\frac{x}{3}+\displaystyle\frac{z}{2}+l-24)a=0\)
\(\displaystyle -x-z\leq{0}\) \(\displaystyle (-x-z)b=0\)
But then I don't know how to get the solutions.
Thanks
\(\displaystyle f(u,z,l)=\sqrt[ ]{(xz)^2+l^2}\)
with the next conditions:
\(\displaystyle \displaystyle\frac{x}{3}+\displaystyle\frac{z}{2}+l=24\)
\(\displaystyle x\geq{0}\);\(\displaystyle z\geq{0}\);\(\displaystyle l>o\)
\(\displaystyle x+z>0\)
I do the next lagrangian:
\(\displaystyle L=-(xz)^2-l^2+a(\displaystyle\frac{x}{3}+\displaystyle\frac{z}{2}+l-24)+b(-x-z)\)
and then I have the KKT conditions:
\(\displaystyle -2xz^2+\displaystyle\frac{a}{3}-b\geq{0}\) \(\displaystyle (-2xz^2+\displaystyle\frac{a}{3}-b)x=0\)
\(\displaystyle -2x^2z+\displaystyle\frac{a}{2}-b\geq{0}\) \(\displaystyle (-2x^2z+\displaystyle\frac{a}{2}-b)z=0\)
\(\displaystyle -2l+a\leq{0}\) \(\displaystyle (-2l+a)l=0\)
\(\displaystyle \displaystyle\frac{x}{3}+\displaystyle\frac{z}{2}+l-24=0\) \(\displaystyle (\displaystyle\frac{x}{3}+\displaystyle\frac{z}{2}+l-24)a=0\)
\(\displaystyle -x-z\leq{0}\) \(\displaystyle (-x-z)b=0\)
But then I don't know how to get the solutions.
Thanks